Find the sum of all the terms in the **nth** row of the series given below.

1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 .......................... ............................ (so on)

**Examples:**

Input : n = 2 Output : 18 terms in 2nd row and their sumsum= (3 + 4 + 5 + 6) = 18 Input : n = 4 Output : 132

**Naive Approach:** Using two loops. Outer loop executes for **i = 1 to n** times. Inner loop executes for **j = 1 to 2 * i** times. Counter variable **k** to keep track of the current term in the series. When **i = n**, the values of **k** are accumulated to the sum.

Time Complexity: O(k), where k is the total number of terms from the beginning till the end of the nth row.

**Efficient Approach:** The sum of all the terms in the **nth** row can be obtained by the formula:

Sum(n) = n * (2 * n^{2}+ 1)

The proof for the formula is given below:

**Prerequisite:**

- Sum of
**n**terms of an Arithmetic Progression series with**a**as the first term and**d**as the common difference is given as:**Sum = (n * [2*a + (n-1)*d]) / 2** - Sum of 1st
**n**natural numbers is given as:**Sum = (n * (n + 1)) / 2**

**Proof:**

Let the number of terms from the beginning till the end of thenthrow bep. Herep= 2 + 4 + 6 + .....nterms For the givenAPseries,a= 2,d= 2. Using the above formula for the sum ofnterms of the AP series, we get,p = n * (n + 1)Similarly, let the number of terms from the beginning till the end of the(n-1)throw beq. Hereq= 2 + 4 + 6 + .....n-1terms For the givenAPseries,a= 2,d= 2. Using the above formula for the sum ofn-1terms of the AP series, we get,q = n * (n - 1)Now, Sum of all the terms in the nth row = sum of 1stpnatural numbers - sum of 1stqnatural numbers =(p * (p + 1)) / 2 - (q * (q + 1)) / 2Substituting the values ofpandqand then solving the equation, we will get,Sum of all the terms in the nth row = n * (2 * n^{2}+ 1)

## C++

`// C++ implementation to find the sum of all the ` `// terms in the nth row of the given series ` `#include <bits/stdc++.h> ` ` ` `using` `namespace` `std; ` ` ` `// function to find the required sum ` `int` `sumOfTermsInNthRow(` `int` `n) ` `{ ` ` ` `// sum = n * (2 * n^2 + 1) ` ` ` `int` `sum = n * (2 * ` `pow` `(n, 2) + 1); ` ` ` `return` `sum; ` `} ` ` ` `// Driver program to test above ` `int` `main() ` `{ ` ` ` `int` `n = 4; ` ` ` `cout << ` `"Sum of all the terms in nth row = "` ` ` `<< sumOfTermsInNthRow(n); ` ` ` `return` `0; ` `} ` |

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## Java

`// Java implementation to find the sum of all the ` `// terms in the nth row of the given series ` ` ` `import` `static` `java.lang.Math.pow; ` ` ` `class` `Test { ` ` ` `// method to find the required sum ` ` ` `static` `int` `sumOfTermsInNthRow(` `int` `n) ` ` ` `{ ` ` ` `// sum = n * (2 * n^2 + 1) ` ` ` `int` `sum = (` `int` `)(n * (` `2` `* pow(n, ` `2` `) + ` `1` `)); ` ` ` `return` `sum; ` ` ` `} ` ` ` ` ` `// Driver method ` ` ` `public` `static` `void` `main(String args[]) ` ` ` `{ ` ` ` `int` `n = ` `4` `; ` ` ` `System.out.println(` `"Sum of all the terms in nth row = "` ` ` `+ sumOfTermsInNthRow(n)); ` ` ` `} ` `} ` |

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## Python3

`# Python 3 implementation to find ` `# the sum of all the terms in the ` `# nth row of the given series ` `from` `math ` `import` `pow` ` ` `# function to find the required sum ` `def` `sumOfTermsInNthRow(n): ` ` ` ` ` `# sum = n * (2 * n^2 + 1) ` ` ` `sum` `=` `n ` `*` `(` `2` `*` `pow` `(n, ` `2` `) ` `+` `1` `) ` ` ` `return` `sum` ` ` `# Driver Code ` `if` `__name__ ` `=` `=` `'__main__'` `: ` ` ` `n ` `=` `4` ` ` `print` `(` `"Sum of all the terms in nth row ="` `, ` ` ` `int` `(sumOfTermsInNthRow(n))) ` ` ` `# This code is contributed ` `# by Surendra_Gangwar ` |

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## C#

`// C# implementation to find the sum of all the ` `// terms in the nth row of the given series ` `using` `System; ` ` ` `class` `Test { ` ` ` `// method to find the required sum ` ` ` `static` `int` `sumOfTermsInNthRow(` `int` `n) ` ` ` `{ ` ` ` `// sum = n * (2 * n^2 + 1) ` ` ` `int` `sum = (` `int` `)(n * (2 * Math.Pow(n, 2) + 1)); ` ` ` `return` `sum; ` ` ` `} ` ` ` ` ` `// Driver method ` ` ` `public` `static` `void` `Main() ` ` ` `{ ` ` ` `int` `n = 4; ` ` ` `Console.Write(` `"Sum of all the terms in nth row = "` ` ` `+ sumOfTermsInNthRow(n)); ` ` ` `} ` `} ` ` ` `// This code is contributed by vt_m. ` |

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## PHP

`<?php ` `// PHP implementation to find ` `// the sum of all the terms in ` `// the nth row of the given series ` ` ` `// function to find the required sum ` `function` `sumOfTermsInNthRow(` `$n` `) ` `{ ` ` ` ` ` `// sum = n * (2 * n^2 + 1) ` ` ` `$sum` `= ` `$n` `* (2 * pow(` `$n` `, 2) + 1); ` ` ` `return` `$sum` `; ` `} ` ` ` ` ` `// Driver Code ` ` ` `$n` `= 4; ` ` ` `echo` `"Sum of all the terms in nth row = "` `, ` ` ` `sumOfTermsInNthRow(` `$n` `); ` ` ` `// This code is contributed by ajit ` `?> ` |

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Output:

Sum of all the terms in nth row = 132

This article is contributed by **Ayush Jauhari**. If you like GeeksforGeeks and would like to contribute, you can also write an article using contribute.geeksforgeeks.org or mail your article to contribute@geeksforgeeks.org. See your article appearing on the GeeksforGeeks main page and help other Geeks.

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